question_answer

The centre of mass of a non-uniform rod of length \[L\] whose mass per unit length \[\lambda \] varies as \[\lambda =\frac{k.{{x}^{2}}}{L}\]where k is a constant and x is the distance of any point on rod from its one end, is (from the same end)

A) \[\frac{3}{4}L\]

B) \[\frac{1}{4}L\]

C) \[\frac{k}{L}\]             

D) \[\frac{3k}{L}\]

Correct Answer: A

Solution :

[a] \[\therefore {{x}_{cm}}\frac{\int\limits_{0}^{L}{\frac{K}{L}{{x}^{2}}dx.x}}{\int\limits_{0}^{L}{\frac{K}{L}{{x}^{2}}dx}}=\frac{{{\left. \frac{{{x}^{4}}}{4} \right|}^{L}}_{0}}{{{\left. \frac{{{x}^{3}}}{3} \right|}^{L}}_{0}}=\frac{3}{4}L\]